What I have:
test
more text
@user653434 text and so
test
more text
@user9659333 text and so
I'd like to filter this text and finally get the following list as .txt file:
user653434
user9659333
It's important to get the names without "@" sign. Thx for help ;)
Using grep -P (requires GNU grep):
$ grep -oP '(?<=@)\w+' File
user653434
user9659333
-o tells grep to print only the match.
-P tells grep to use Perl-style regular expressions.
(?<=@) tells sed that @ must precede the match but the @ is not included in the match.
\w+ matches one or more word characters. This is what grep will print.
To change the file in place with grep:
grep -oP '(?<=@)\w+' File >tmp && mv tmp File
$ sed -En 's/^@([[:alnum:]]+).*/\1/p' File
user653434
user9659333
And, to change the file in place:
sed -En -i.bak 's/^@([[:alnum:]]+).*/\1/p' File
-E tells sed to use the extended form of regular expressions. This reduces the need to use escapes.
-n tells sed not to print anything unless we explicitly ask it to.
-i.bak tells sed to change the file in place while leaving a backup file with the extension .bak.
The leading s in s/^@([[:alnum:]]+).*/\1/p tells sed that we are using a substitute command. The command has the typical form s/old/new/ where old is a regular expression and sed replaces old with new. The trailing p is an option to the substitute command: the p tells sed to print the resulting line.
In our case, the old part is ^@([[:alnum:]]+).*. Starting from the beginning of the line, ^, this matches @ followed by one or more alphanumeric characters, ([[:alnum:]]+), followed by anything at all, .*. Because the alphanumeric characters are placed in parens, this is saved as a group, denoted \1.
The new part of the substitute command is just \1, the alphanumeric characters from above which comprise the user name.
Here, the s indicates that we are using a sed substitute command. The usual form
With GNU grep:
grep -Po '^@\K[^ ]*' file
Output:
user653434 user9659333
